# Notes for Section 6_2 Setting up integral.pdf - Sec 6.2...

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Sec. 6.2: Setting Up Integrals: Volume, Density, Average Value The methods used in Section 6.1 to compute area by integration can be modified to compute the volume of a solid region! We will develop methods for computing volume of several kinds of solids in this section. We are also going to use integral to compute other quantities such as total mass and average value of a function. 1. Volume – The Cross-Sectional (or Slicing) Method The slicing method can often be used to find the volume of a solid that can be sliced up into parallel slabs whose faces have easily computed areas. To this end, let us consider Figure 1 below. Figure 1 Let V be the volume of this solid body and h be the height along the y -axis. The intersection of the solid with the horizontal plane (or slice) at height y is called the horizontal cross section at height y . We are going to call the area of the cross section at height y A ( y )’. (Note that in general, the area of each cross section depends on the height y . For this particular figure, higher from the base you go, the smaller the cross section it gets.) What we are going to do is to slice up this solid “horizontally” into N pieces, which is analogous to subdividing an interval [ a, b ] into N rectangles in Section 5.1. Let us pick one sample slice, call it the ‘ i th slice’ such that the base of this slice is ‘ y i - 1 ’ from the base of the original solid and whose height (or thickness) is ‘Δ y ’. Also, let Δ V be the volume of this slice. Observe that the base area of this particular slice is given by A ( y i - 1 ). Observe further that if N is sufficiently large, then the thickness Δ y is very small and the slices are very thin. In this case, the i th slice is nearly a right cylinder of base area A ( y i - 1 ) and height Δ y . Thus its volume, namely Δ V , can be computed as follows. Δ V A ( y i - 1 ) Δ y. (1) Then the approximate total volume of this solid can be expressed in terms of the following Riemann sums. V N X i =1 A ( y i - 1 ) Δ y. (2) Taking the limit of Equation (2) as Δ y 0 (or equivalently, N → ∞ ), we obtained the exact volume as the definite integral.